Given that a is a positive constant and
\[ \int_{a}^{2a} \frac{t+1}{t} \, dt = \ln 7 \]
show that \( a = \ln k \), where k is a constant to be found. - Edexcel - A-Level Maths Pure - Question 5 - 2017 - Paper 1
Question 5
Given that a is a positive constant and
\[ \int_{a}^{2a} \frac{t+1}{t} \, dt = \ln 7 \]
show that \( a = \ln k \), where k is a constant to be found.
Worked Solution & Example Answer:Given that a is a positive constant and
\[ \int_{a}^{2a} \frac{t+1}{t} \, dt = \ln 7 \]
show that \( a = \ln k \), where k is a constant to be found. - Edexcel - A-Level Maths Pure - Question 5 - 2017 - Paper 1
Step 1
Write \( \int_{a}^{2a} \frac{t+1}{t} \, dt \) in simpler terms.
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Answer
First, separate the integrand:
[ \int \frac{t+1}{t} , dt = \int \left(1 + \frac{1}{t} \right) , dt = \int 1 , dt + \int \frac{1}{t} , dt = t + \ln |t| + C ]
Then, evaluate the definite integral from a to 2a:
[ \int_{a}^{2a} \frac{t+1}{t} , dt = \left[ t + \ln |t| \right]_{a}^{2a} = (2a + \ln(2a)) - (a + \ln(a)) ]
Step 2
Show that \( (2a + \ln(2a)) - (a + \ln(a)) = \ln 7 \)
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Answer
Simplifying this yields:
[ (2a + \ln(2) + \ln(a)) - (a + \ln(a)) = (2a - a) + \ln(2) = a + \ln(2) ]
Thus, we have [ a + \ln(2) = \ln 7 ]
Step 3
Solve for \( a \) in terms of \( k \).
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Answer
Rearranging gives us:
[ a = \ln(7) - \ln(2) = \ln \left( \frac{7}{2} \right) ]
This means ( a = \ln k ) where ( k = \frac{7}{2} ).
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